## Facebook Interview Question for Software Engineer / Developers

Country: United States
Interview Type: Phone Interview

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7
of 7 vote

Events that should happen are (i) dice A is not 6, (ii) dice B is not six, both at the same time, both events are independent. Hence P = P_A*P_B = 5/6*5/6=25/36~= 0.69

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0

that's correct !!

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0
of 0 vote

P = (1-13/36) ~= 0.64

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0

Sorry 0.69

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0
of 0 vote

(5/6)^2

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0
of 2 vote

there are total 5 possible ways to get Count 6 on two Dice
(1,5),(2,4),(3,3),(4,2),(5,1) = 5 possible ways
total possibilities on Two Dice is 36 so it should be 5/36 = 0.13

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0
of 0 vote

possiblity for not 6 come on dice A is 5/6
similarly ,
possiblity for not 6 come on dice B is 5/6
for both having no 6 is p(A)*p(B) = 5*5/6*6 = > 25/36

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0
of 0 vote

@autoboli is correct and has the shortest solution possible, but if you want to go the longer way we can also do this:
Proability of getting no sixes is 1 - P_getting_1_six - P_getting_two_sixes, where P_getting_1_six = 2 * (1/6 * 5/6) = 10/36 and P_getting_two_sixes = 1/6 * 1/6 = 1/36. 1 - 10/36 - 1/36 = 1 - 11/36 = 25/36 = 0.69. I know it's a long way but just showing another way to do it :)

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0
of 0 vote

1.
What is the set of all possible outcomes when I roll a dice?

``````d1 = dice 1.
If dice 1 has the value in column 1, then what are the possible sums of both dices?

d1    Outcomes possible
1	2 3 4 5 6 7
2	3 4 5 6 7 8
3	4 5 6 7 8 9
4	5 6 7 8 9 10
5	6 7 8 9 10 11
6	7 8 9 10 11 12``````

2.
Total outcomes = 36
Prob of getting 6 = 5/36
Prob of not getting 6 = 31/36.

This is not the same as the others...so, any help is appreciated. Thanks!

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1
of 1 vote

I see what you're trying to do here, and there is a way to make it work out. Essentially, using your setup, the way you would remove all rolls that include a 6 would be to remove the bottom row *and* the rightmost column, as each of those represent any pair of rolls that contain a 6. What you're then left with is a 5x5 matrix, which is 25 possibilities out of your original 36.

``````+	1 2 3 4 5   | 6

1	2 3 4 5 6   | 7
2	3 4 5 6 7   | 8
3	4 5 6 7 8   | 9
4	5 6 7 8 9   | 10
5	6 7 8 9 10  | 11
--------------------
6	7 8 9 10 11 | 12``````

25 / 36 = ~0.69

The more straightforward way to think of it is to consider the dice entirely independent of each other, because the second roll does not depend on the first. So you need only consider the probabilities of each, multiplied together. That is, the probability of not rolling a 6 with each die is 5/6. Then you have:

5/6 * 5/6 = 25/36 = ~.69

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1
of 1 vote

Super thanks Nick! I realized my mistake. I misinterpreted the question as the sum of the two dices should be 6...the question is the non-occurrence of any 6 in either of the two dices. As soon as I read your answer, I realized my mistake. So, thanks much!

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0
of 0 vote

25/36

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0
of 0 vote

P(A = 6) = 1/6
P(B = 6) = 1/6

``````P(A ==6 U B == 6) = 1/6 + 1/6 - (1/6*1/6)

Now to get the prob of not getting any 6 =
1 - P(A ==6 or B == 6)
= 1-2/6-1/36 = 25/36 = 0.69``````

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